The maximum area of a hexagon that can inscribed in an ellipse I have to find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as $\frac{x^2}{16}+\frac{y^2}{9}=1$. 
I considered the circle with the major axis of the ellipse as the diameter of the circle with centre as the origin. By considering lines of $y=\pm\sqrt{3}x$, you can find the points on the circle which are vertices of a regular hexagon(including the end points of the diameter). I brought these points on the given ellipse such that the $x$ coordinate of the vertices remain the same and only the $y$ coordinate changes depending upon the equation of the ellipse. But that did not lead to a hexagon. 
Since the question nowhere mentions of the hexagon being regular, I think it might be a irregular hexagon but any such hexagon can be drawn which encompasses most of the area of the ellipse. 
 A: We can use the affine transformation that $x=4u,y=3v$. Therefore, The ellipse will maps to the unit circle $u^2+v^2=1$. One of the most important property of affine thansformations is invariance of ratio of the areas. Ratio of areas of ellipse and unit circle is $4\cdot 3 = 12$.Now we have to find a cyclic hexagon that has maximum area. (By isoperimetric inequaliy, this hexagon must be regular). 
Area of regular hexagon is $\dfrac{3\sqrt3}{2}$ and therefore, the hexagon has maximum area that in ellipse is $$12\cdot \dfrac{3\sqrt3}{2} =18\sqrt3 $$
A: 
A visual clue. The largest hexagon in a circle is regular. Stretch this to obtain the hexagon in the ellipse.
A: Consider the standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. An hexagon cab be inscribed in this ellipse whose area is 4 times the area of a rectangle plus a right angled triangle as
$$A_H(t)=4[ab\sin t \cos t+\frac{1}{2} b \sin t (a-a\cos t)]=ab[\sin 2t+2\sin t]~~~~~(1)$$ where $P(a \cos t, b \sin t)$ is the point on the ellipse and let the vertex be $A(a,0)$ in the first quadrant. Now let us maximize $A_H(t)$, where $\frac{dA_H}{dt}=0$ gives $2\cos 2t+2\cos t=0 \implies  t=\pi/3, \pi$. By choosing $t=\pi/3$, we get $A''_H(\pi/3)<0.$
Finally, $max(A_H)=A_H(\pi/3)=\frac{3\sqrt{3}}{2}ab=18\sqrt{3}$ (when $a=4, b=3).$
